Throughout this post is a field. Recall that if is a -vector space, then the dual space of is the vetor spcae consisting of all -linear maps from to If is finite dimensiona, say with the basis then for every we define by and for all It is easy to see that is a basis for In particular,

**Theorem**. Let be a finite dimensional -algebra. Then is a Frobenius -algebra if and only if there exists such that does not contain any non-zero left ideal of

*Proof*. Suppose first that is Frobenius. So there exists a non-degenerate bilinear form of such that for all Define by for all Obviously because is linear. Now suppose that is a left ideal of and Let and Then and so Therefore because is non-degenerate. So we have proved that Conversely, suppose that there exists such that does not contain any non-zero left ideal of Define by for all Clearly is bilinear because is linear. Also, clearly for all To complete the proof of the theorem we only need to show that is non-degenerate. So suppose that for some and all Then and thus for all i.e. Hence because is a left ideal of and does not contain any non-zero left ideal of So and we are done.

**Example 1**. If is a finite field extension, then is a Frobenius -algebra.

*Proof*. Let be a -basis for Define by for all Clearly is a non-zero -linear map and does not contain any non-zero ideal of because is a field and so it has no non-zero proper ideal.

**Example 2**. Let be a finite group. Then the group algebra is a Frobenius -algebra.

*Proof*. Let where is the identity element of Define the map by for all Clearly is a -linear map and Suppose that and is a non-zero left ideal of Let Then for some Since is a left ideal of we have But the coefficient of in is and so which is a contradiction. Thus